Answer
77.1 grams
Work Step by Step
$q_{Ag}=-q_{water}$
$\implies m_{Ag}\times c_{Ag}\times\Delta T_{Ag}=-m_{water}\times q_{water}\times\Delta T_{water}$
$\implies m_{Ag}=-\frac{m_{water}\times c_{water}\times\Delta T_{water}}{c_{Ag}\times\Delta T_{Ag}}$
$=-\frac{(100.0\,g)(4.18\,J/g\cdot ^{\circ}C)(26.2^{\circ}C-24.8^{\circ}C)}{(0.235\,J/g \cdot ^{\circ}C)(26.2^{\circ}C-58.5^{\circ}C) }$
$=77.1\,g$
